Problem: Divide the following complex numbers. $ \dfrac{-21+20i}{-5+2i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-5-2i}$ $ \dfrac{-21+20i}{-5+2i} = \dfrac{-21+20i}{-5+2i} \cdot \dfrac{{-5-2i}}{{-5-2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-21+20i) \cdot (-5-2i)} {(-5+2i) \cdot (-5-2i)} = \dfrac{(-21+20i) \cdot (-5-2i)} {(-5)^2 - (2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-21+20i) \cdot (-5-2i)} {(-5)^2 - (2i)^2} = $ $ \dfrac{(-21+20i) \cdot (-5-2i)} {25 + 4} = $ $ \dfrac{(-21+20i) \cdot (-5-2i)} {29} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-21+20i}) \cdot ({-5-2i})} {29} = $ $ \dfrac{{-21} \cdot {(-5)} + {20} \cdot {(-5) i} + {-21} \cdot {-2 i} + {20} \cdot {-2 i^2}} {29} $ Evaluate each product of two numbers. $ \dfrac{105 - 100i + 42i - 40 i^2} {29} $ Finally, simplify the fraction. $ \dfrac{105 - 100i + 42i + 40} {29} = \dfrac{145 - 58i} {29} = 5-2i $